交换numpy数组的尺寸

Question

I would like to do the following:

for i in dimension1:
  for j in dimension2:
    for k in dimension3:
      for l in dimension4:
        B[k,l,i,j] = A[i,j,k,l]

without the use of loops. In the end both A and B contain the same information but indexed differently.

I must point out that the dimension 1,2,3 and 4 can be the same or different. So a numpy.reshape() seems difficult.

Answer 1

The canonical way of doing this in numpy would be to use np.transpose 's optional permutation argument. In your case, to go from ijkl to klij , the permutation is (2, 3, 0, 1) , eg:

In [16]: a = np.empty((2, 3, 4, 5))

In [17]: b = np.transpose(a, (2, 3, 0, 1))

In [18]: b.shape
Out[18]: (4, 5, 2, 3)

Answer 2

Please note: Jaime's answer is better. NumPy provides np.transpose precisely for this purpose.

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Or use np.einsum ; this is perhaps a perversion of its intended purpose, but the syntax is quite nice:

In [195]: A = np.random.random((2,4,3,5))

In [196]: B = np.einsum('klij->ijkl', A)

In [197]: A.shape
Out[197]: (2, 4, 3, 5)

In [198]: B.shape
Out[198]: (3, 5, 2, 4)

In [199]: import itertools as IT    
In [200]: all(B[k,l,i,j] == A[i,j,k,l] for i,j,k,l in IT.product(*map(range, A.shape)))
Out[200]: True

Answer 3

You could rollaxis twice:

>>> A = np.random.random((2,4,3,5))
>>> B = np.rollaxis(np.rollaxis(A, 2), 3, 1)
>>> A.shape
(2, 4, 3, 5)
>>> B.shape
(3, 5, 2, 4)
>>> from itertools import product
>>> all(B[k,l,i,j] == A[i,j,k,l] for i,j,k,l in product(*map(range, A.shape)))
True

or maybe swapaxes twice is easier to follow:

>>> A = np.random.random((2,4,3,5))
>>> C = A.swapaxes(0, 2).swapaxes(1,3)
>>> C.shape
(3, 5, 2, 4)
>>> all(C[k,l,i,j] == A[i,j,k,l] for i,j,k,l in product(*map(range, A.shape)))
True

Answer 4

I would look at numpy.ndarray.shape and itertools.product:

import numpy, itertools
A = numpy.ones((10,10,10,10))
B = numpy.zeros((10,10,10,10))

for i, j, k, l in itertools.product(*map(xrange, A.shape)):
    B[k,l,i,j] = A[i,j,k,l]

By "without the use of loops" I'm assuming you mean "without the use of nested loops", of course. Unless there's some numpy built-in that does this, I think this is your best bet.

Answer 5

One can also leverage numpy.moveaxis() for moving the required axes to desired locations. Here is an illustration, stealing the example from Jaime's answer :

In [160]: a = np.empty((2, 3, 4, 5))

# move the axes that are originally at positions [0, 1] to [2, 3]
In [161]: np.moveaxis(a, [0, 1], [2, 3]).shape 
Out[161]: (4, 5, 2, 3)